Tuesday, 5 August 2014

Summary: Genetics, math and statistics

Variance components

Variance and variance analysis have been discussed earlier in this blog, so for now let's just remind ourselves that variance expresses the variation of the data: how far apart are the extreme values in the current dataset. Variance is also expressed in the same units than the data, so the unit affects the amount of variance (for example 1,5kg vs 1500g). 

Variance components are just the different factors that create variation between measurements. 
Variance components in animal breeding can be viewed from two perspectives: genetic vs environmental contribution (which together are the phenotypic variance), or dam/sire vs residual variance. Additive, maternal and dominance effects together form the genetic contribution. Common and general environment form the environmental contribution . Therefore

P = Var(A) + Var(M) + Var(D) + Var(e) + Var(eg).

All these contributions are built from variance from the dam, the sire and the residual. The chart below shows how the different components (pillars) are built. Note that sire variance (Var(s)) affects only additive genetic variance, of which it constitutes 25 %. Therefore Var(s) = 0,25 Var(A), and Var(A) = 4*Var(s). Dam variance is also 25 % of the additive genetic effect and dominance effect, but 100 % of common environmental and maternal effects.

Only some of the components above are inherited from parent to offspring. The rest are often ignored, so the formula can be simplified into

Var(P) = Var(A) + Var(M) + Var(e).

One interesting aspect about variation is that when the reliability of the breeding value, rTI, increases, the variance of estimated breeding values increases as well. This is because the higher the reliability, the better we see the differences between the animals, and the more variation we get. However, when the reliability increases, the variance of the true breeding values decreases between animals with the same estimated breeding value. This is of course because the increased reliability brings our estimation closer to the true breeding value. One has to consider variance in its context.

Heritability h2 is often written as additive genetic variance divided by phenotypic variance, i.e Var(A) / Var(P). Considering the previous formulas, heritability can be deduced from sire variance:  h2 = 4 * (Var(s) / Var(P)).


Change of gene frequency under selection is

where q1 is the frequency of the selected gene after one generation, q is the square root of the original frequency (q2), s is the coefficient of selection and q2 is the original frequency as per the Hardy-Weinberg equation (q2 + 2pq + p2 = 1). 

Number of generations required
The number of generations required to achieve a certain breeding objective is calculated thusly:
where t is the number of generations, qt is the gene frequency after t generations and q0 is the original gene frequency. qt = q0 / (1 + tq0).


Coefficient of selection, s
Coefficient of selection is the proportionate reduction of gametic contribution of a genotype compared to the standard genotype. It shows how much less animals of a certain genotype, usually the less facorable, affect the next generation when selection takes place. For example how much less gametes do unpolled animals contribute compared to polled, when polled ones are selected for breeding.
The contribution of the favorable genotype is 1, the coefficient is s so the contribution of the less favorable genotype is 1-s.
If s = 0,1, then the contribution of the favorable genotype is 1 and the contribution (and the fitness) of the less favorable is 1-0,1 = 0.9. In practice, for each 100 zygotes by the favorable genotype, 90 zygotes are born by the less favorable.

Tuesday, 29 April 2014

Feeding for healthy dairy cattle

In this post we begin by explaining the concept of feeding strategy. Two basic strategies, total mixed ratio and partial mixed ratio, are discussed. We then move on to feeding heifers and lactating cows, with extra attention being paid calving time. Lastly we review some aspects about feeding and cow health.

A feeding strategy is a combination of the feeds used and the machinery and methods used to prepare and distribute the feed. Selection of a feeding strategy depends on
  • The genetic potential of the cattle
  • Production targets, possible production quotas
  • Location of the farm (available fields, distance from fields and distance from other sources of cheap feed, i.e. bakeries and breweries)
  • Health and welfare of the animals
  • Herd size
  • Buildings and machinery available for use
  • Number and skills of workers available
  •  Available feeds and their prices (connected to the location of the farm)
  • Digestibility and quality of the feeds grown in-farm
  • Production type (normal / organic / biodynamic)
In short, the two different feed distribution techniques for dairy cattle are total mixed ratio (TMR) or partial mixed ratio (PMR). In TMR, each ingredient is added to a single mix, which is then distributed to all animals. In PMR silage and concentrate are fed separately, while minerals and vitamins may be mixed into either one (or fed separately). In Northern Europe TMR has become increasingly popular, although it is difficult if there are several animal groups in one farm. For specialized farms TMR is an excellent choice.

Several studies show that TMR and PMR are equally effective in terms of milk yield. With TMR the increase of eating potential after calving is more stable than with PMR, but the total intake of feed is not affected by the feeding strategy. TMR may decrease the changes of rumen pH during changes in feeding. Altering the TMR may not be necessary during lactation, because cows in late lactation do not gain excess weight even when fed with the same mic than cows in the early lactation. Number of feeding times in a day does not affect milk yield, but feeding only once decreases the utilization of the feed. TMR may also decrease illnesses throughout the lactation (perhaps due to the stabilizing effect on rumen pH).


A good dairy cow is the result of a successful heifer rearing. Today, most female calves are needed as replacements due to high percentage of culling and young age of the culled animals. Rearing calves and heifers is expensive, but vital for maintaining the productivity of the farm.

From birth to 3 months, when the udder tissue develops isometrically, female calves are fed ad libitum
 Feeding ad libitum helps the calf to utilize her full genetic potential. From 3 moths until 10-12 months the udder develops faster than other tissues, and strong feeding disrupts that growth by increasing fat tissue and decreasing the amount of secreting tissue. High-energy diet for heifers may decrease their milk yield up to 52 %, but the scientific results are not unambiguous. It is recommended that Holstein heifers shouldn't grow more than 650-900 g/day and Ayrshires 600-800 g/day. Slow growth is not harmful, but increases costs and offsets the optimal time for 1st insemination (15 mo) and calving (24 mo).
Heifers of 3-12 mo of age can be fed with straw, hay, grass, silage, or kept on pasture. Silage with a high D-value should be avoided, and concentrate (max 2 kg/day) should be given only if needed. Weight of the heifers should be measured for example by measuring the girth of their chest. Feeding affects the weight and the width of the hip of the heifers, but does not have impact on their height at withers or the width of the body. To ensure healthiness, minerals and vitamins must be made available. Once the heifer reaches sexual maturity, the growth of the udder decreases. Sexually mature heifers can be fed more freely without adverse effects on their growth or production.

The correct time for insemination is determined by the breed and weight of the heifer. Large breeds should be allowed to grow larger prior to insemination. Otherwise their 1st lactation will be unproductive, because most of the energy is needed for the animal's own growth instead of milk production. Calvings may also be problematic if the dam is still too small. Ayrshires should be at least 14 months of age at the insemination, weigh over 320 kilos and have a chest girth over 158 cm. The same values for Holsteins are 15 mo, over 340 kg and over 162 cm. As a comparison, for a Finnish landrace the target weight is > 240 kg and chest girth 140 cm due to the small adult size of the breed. Note that a heifer is not in its adult weight until at 5 years of age - after 3 calvings!

Recommended daily grow for heifers
(data from several sources).
Feeding a heifer in gestation aims at allowing the heifer and her calf to grow normally without the dam getting too fat before calving. Feeding is limited during the first 6 months of gestation. Daily the animals should be given a limited amount of silage, 0,5 - 2 kg of concentrate plus minerals and vitamins. During the last 3 months ME MJ per day is increased 11-34 ME MJ /day to ensure the correct body condition score (BCS) during calving. If the BCS is over 3.5, the risk for metabolic illnesses and drastic weight loss after calving increases. In a study by Mäntysaari et al (1999), best milk yield was achieved when the heifers grew 650 g/day during the firsth 6 months and 850 g/day during the last three.

To summarize, heifers should be fed in different phases. From 3 - 12 months and during the first six months in gestation the feeding should be limited. From 12 months to fertilization the heifers are fed to ensure the correct weight on insemination. The last months of gestation the heifers are again fed to support their healthy growth and a correct BCS. During the last weeks of gestation, the heifer is introduced to the feeds and concentrates of cows to get her ready for her first lactation.

Measuring a heifer.
(c) Land O'Lakes, Inc.



There are several strategies for dairy cattle. In feeding per norms there are two options: individual feeding per norms, where the amount of concentrate is tailored for each animal based on their last measured milk yield (may be automatic if milking robots and concentrate kiosks are used). In an applied feeding per norms -strategy the amount of concentrate is increased during the production peak, and then returned to a normal level. In both strategies silage is freely available, but the amount of concentrate is based on a predicted or measured milk yield. Equal concentrate -strategy means that each cow gets the same amount of concentrate during a certain production stage. Silage is freely available, so the cows use it to adjust their energy intake. At the end of lactation each cow may be fed individually as per their current BCS. Equal concentrate can be used for each individual animal, or by tailoring the concentrate amounts per herds or animal groups (dry cows, pregnant heifers etc). Both strategies are equally effective.

Note that when fed freely, there must be 10-15 % of the feed left over! If the animals eat everything, it is a clear indication that they are not receiving enough feed. Leftovers can be then fed to weaned calves, heifers, growing bulls or to dry cows, who in general are less picky than lactating cows. High-quality silage can be offered even to unweaned calves to get them aquainted with the taste and feel of "real food".

Addition of concentrate has different production responses depending on the phase of lactation. In early lactation increasing the amount of concentrate increases milk yield more than in the late lactation, but only during the early phase. If the portion of concetrate is increased quickly, milk yield can be increased up to 3 kg / day during the first 5 weeks (Kokkonen 2004). There is no impact to the milk yield in mid or late lactation.

(c) Wikipedia Commons
The increment lowers the amount of silage eaten approximately 0,4 - 0,6 kg for each added kg of concentrate. Because the production peak occurs in the early lactation, in the middle lactation most of the energy from the concentrate is used for the damn's own growth. Effects of increasing the amount of protein gained from the diet is not dependent on the lactation phase. However, inadequate protein percentage in early lactation does lower milk yield noticeably.

After calving cows utilize their own tissues to produce enough nutritients to their milk. This leads to weight loss and a decrease in BCS. The more heavily the cows were fed during early gestation the more weight they lose in early lactation. Weight loss is also connected to limited feeding during early lactation. Due to effective tissue mobilization there is no decrease in milk yield even if the amount of concentrate is low. However, these cows may yield less milk during the entire lactation, and have lowered fertility after the lactation. Therefore it is advised to offer at least enough concentrate during the early lactation.

Feeding and calving

Calving begins a major change in the nutritional requirements of the body. The udder requires three times more glucose, twice more amino acids and 4-5 times more fatty acids than the developing embryo did. If the pregnant cows are fed heavily before calving, they suffer from many adverse effects during early lactation. There are more non-esterized fatty acids (NEFA) in their bloodstream, which indicates a high level of tissue mobilization. During the first month of lactation, 40-50 % of the fats in the milk are tissue-based.

When there's a large amount of NEFAs in the blood, the animal's body cannot effectively use them all. Fats begin to accrue in the liver, which again hinders the functions of the liver even further. This may lead to fatty liver. A fatty liver cannot produce glucose as much as the cow needs (glucose is the main source of lactose in the milk). To replace the glucose the cow needs, she again utilizes her body and uses fats as an energy source. This leads to weight loss and ketosis, since using fats in high amounts as energy produces harmful metabolites.

Both a fatty liver and ketosis are caused by metabolic stress. Metabolic stress is natural after parturition in all mammals, but due to breeding, the stress is harmfully high for dairy cows. The stress decreases fertility and immunity, making the animal more vulnerable to opportunistic pathogens and diseases such as mastitis. Diseases caused by metabolic stress are the main cause for culling cows. One could say that we have, while selecting cows based on their high productivity, bred cows for fast culling. It should be crystal clear that breeding and animal management must focus on increasing the health and endurance of cows. Luckily in some areas, like in the Scandinavia, this has been the aim of cow breeding for decades.

Breeding alone cannot do much to prevent metabolic stress without decreasing milk yield considerably. Faster and better options are to
  • Keep the BCS of cows at 3 - 3,5 before calving
  • Provide enough high-quality silage after calving
  •  Provide enough cereal-based concentrate in the early lactation
  • Minimize all stress before calving, ensure the cow is in a safe and clean pen
  • Provide enough protein before and after calving (120 - 130 g/ kg DM) to reduce the need to draw calcium from the bones and other tissues. In maize-based feeding protein addition is not necessary.
  • Alternatively, reduce protein prior to calving to "teach" the cow's body to utilize its own Ca effectively, and provide enough Ca after calving to replace the losses.


Maintenance of health and production

Once a farmer has achieved a good level of feeding costs versus income from milk yield, it is time to maintain that situation or even improve it. The overall situation of the herd can be estimated from various indicators.

Comparing the yearly milk yield of the entire herd or single cows to local averages is a good way to see how the farm measures against its "competitors". Annual amounts of milk, protein and fat produced also show the development of the farm. Lifetime production is also a good measure for endurance of the cows. Different cost structures should also be analyzed annually. Comparisons should not be made to farms from others countries, unless you are certain the results are calculated similarly.

Technology offers both invasive and non-invasive methods for monitoring cow's health. Rumen pH can be measured with an automated bolus (by Smaxtec) installed in the rumen, and ruminating can be measured with a simple noseband by RumiWatch. While the bolus may sound dubious, it does not affect the functions of the rumen at all. In fact, after slaughter, items such as scarves, wallets, hats and even a hammer have been found from the rumen of an old cow!

Milk properties can also give hints about the cow's health. Milking robots can measure somatic cell count, conductivity and hormone levels from the milk. Urea is another component which can be measured. It is a by-product of normal protein metabolism, and is present in milk and in blood by the same amounts. The analysis of urea is usually done in the dairy or a laboratory, because specific devices are needed. Urea can be used to estimate the levels of proteins in the feed: if the urea content is high (> 40 mg/100 ml) , the animal cannot utilize all of the nitrogen it receives from feed. If the content is low (< 20 mg/100 ml), added amino acids may be necessary.

All this information should be used wisely, but not rely solely on it. One should also remember that the point is not to rear cows at the razor's edge between highest possible production and metabolic illness!

Monday, 28 April 2014

Predicting and improving dry matter intake of dairy cows

(c) Agefotostock
Feeding cows must consider two aspects: what feeds can be grown in the farm and what needs to be bought, and how the feeding should be optimized. Optimization is either searching for the minimum or the maximum. Optimization problems must consider the factor to be minimized or maximized, the factor to be optimized and its relation to the first factor and different constraints. Different optimization possibilities are for example lowest cost, highest milk yield / protein yield / milk fat yield or minimizing P / N emissions and excess nutrients.

Optimization should also be done for different time periods. Short-term optimization is usually just optimizing feeds based on their current costs and perhaps balancing the homegrown feeds with bought feedstuffs. Medium-term frame optimization is about planning how to use pastures, what crops to grow and how the crop rotation is implemented. Long-term optimization is about what type of feed to use (total mixed ratio, liquid / solid / semi-solid feeds) and what kind of machinery and knowledge is needed to produce and distribute the feed for the animals.

There are several different recommendations for cattle, many of which have been referenced to in this blog. However, every cow, heifer and calf is unique, and every cattle has its own specific needs. Feed optimization should consider the prices of the feeds, the professionally analysed qualities of the home-grown feed and the predicted eating capacity of the animals (TDMI, total dry matter intake). These provide data for calculating the estimated milk yield, which is the basis of calculating the nutritional needs of each animal group. For detailed information about the TDMI, please refer to Huhtanen et al. (2008).TDMI index is calculated for silage DM intake (SDMI) and concetrate DM intake (CDMI) separately. TDMI = SDMI + CDMI - 1000, and each TDMI point is approximately 90-100 g DM / d.

Online SDMI calculator by MTT
SDMI and TDMI are normalized so that an average silage gets 1000 points. The "average" in this system refers to a grass-based silage with 25 % DM, D-value of 680, 80 g/kg of acids and an NDF content of 550 g/kg DM. 1 If a feed gets 100 points, the cows will eat 10 kg DM of the said feed in a day, but only 8 kg DM of a feed with 80 points.  To get a better understanding of TDMI and its components, please refer to the online calculators provided by the Finnish research institute MTT: SDMI calculator and  CDMI / TDMI calculator.

CDMI differs from TDMI and SDMI. It describes how much less silage the animal eats when the amount of concentrate is increased. Generally, the more concentrate is offered, the less silage the animals eat. The replacement ratio is approximately 0.47: for each 1 kg of concentrate added, the cows decrease their consumption of silage by 0,47 kg. Protein in the concentrate impacts consumption to both directions: when the crude protein content of the concentrate incerases over 170 g/kg DM, consumption of feed increases; when the amount of rumen-digestable protein increases, consumption decreases. Cows are hence able to regulate their intake of nutrients by altering their consumption of concentrate and silage (Kuoppala et al 2008).

Several factors affect the eating potential of the cow. The factors can be classified into four main categories:
  • Feed-based factors
  • Animal-based factors
  • Environmental and management factors
  • Interactions.
Examples of changes in the body after feeding.
Feed factors are the amount and quality of feeds used, and their effects in the digestive tract (e.g. metabolites). For example, ammonia from formed from excess amino acids send inhibitory signals to the metabolism. Feed with low digestibility sends physical inhibitory signals as the rumen fills up. Animals factors are size of the cow, milk yield, genetic traits and the state of lactation and age of the animal. Environmental factors consider the management of the animals: how the feed is distributed, number of feeding places in the barn, length of day, climate and weather. Interactions are different interactions between the other factors. For example, some breeds react more strongly to hot weather, and thus eat and produce less.

So why is it important for a cow to eat as much as possible? Dairy cows have been bred to produce spectacular amounts of milk, which far exceeds the needs of the calf. The production of milk requires huge amounts of nutrients, but the cow can only eat a certain amount of feed a day. The nutritients it doesn't receive from feed are absorbed from the animal's own tissues, leading to weight-loss and possible metabolism-related illnesses. Feed optimization can ensure that the cow gets the right amount of right feeds with accetable costs. Increased eating also increases milk yield. Studies have shown that the regression between TDMI and milk yield is 0,9499.

The farmer has several possibilities to increase TDMI and thus increase the milk yield:
  • Ensure a high D-value for the silage. 10g / kg DM increase in D-value increases cfeed consumption by 175 g DM/day.
  • Ensure successful preservation (fermentation, baling, wilting). Oxygen-free preservation and low pH prevent molding and wrong types of fermentation. When the fermentation acids increase 10 g / kg DM, consumption decreases 128 g DM d.
  • Include whole crop silage or forage and cereal silage into the basic grass-based silage. Inclusion of 30-70 % of red clover increases consumption by 1,3 kg and milk yield 1,3 kg / day.
  • Include peas, oat or broad beans to the silage.
  • Optimize the DM content of the silage. Consumption increases linearly until 42 % of DM, and then begins to decrease.

Sunday, 27 April 2014

Feeding dairy cattle

No matter what types of silage, roughage, concentrate and other feeds are used, dairy cows always need a minimum amount of fibre. Fibre is essential for the functions of the rumen, and a well-functioning rumen is the basis of a healthy, well-producing dairy cow. In a daily feed portion, 25-27 % of the dry matter should come from NDF-fibre in the roughage. Fibre not only affects the layering inside the rumen, but also increases ruminating and the amount of saliva excreted. Saliva keeps the pH of the rumen steady, and prevents acidosis.

Not all types of NDF-fibre are as effective. The particle size should be over 3 mm, but increasing the size over 6-10 mm doesn't increase ruminating any further. Fibre from roughage is more efficient than fibre in concentrate, mostly due to the difference in particle size. However, roughage with a high D-value can be low in fibre.Composition of the roughage is also important: legumes have less fibre than hay plants, and their NDF has a lower digestibility, even though the D-value is similar. Legumes and corn have about 150 g less NDF in a 1 kilogram of dry matter.

Crude protein
Dairy cows should get 130-140 g / 1 kg DM of crude protein (CP) daily. CP content higher than 140 g/kg DM is not utilized effectively, and increases N excreted in faeces. Studies show that cattle utilizes 0 % of  the nitrogen, when it added by using fertilizers high in N. In contrast, cows utilise 16 % of nitrogen in silage harvested early. Succesful fermentation and high D-value increase the CP value.

D-value describes the digestible portion of the organic content of the feed. Target D-value in hay-based feed is 650-690 (65-69 %). The earlier the silage is harvested, the higher the D-value is. When the D-value of a feed is high, adding concentrate increases milk yield only slightly. On the other hand, the impacts of poor D-value can only partially be covered by adding concentrate. Low D-value leads to low amounts of feed eaten, which lowers the CP, fibre and other nutrients received.

From the start of June, the D-value of hay in the first harvest decreases 0,5 % a day due to lignification.In the 2nd and 3rd harvests the decrease is much slower, but the feed values and amount of digestible fibre of late harvests are lower anyway. 

Feed ingredients

Italian ryegrass as more leaves and is more palatable than Westerwold ryegrass. The D-value of the Italian variety also decreases slower, but the plant grows more slowly and has more weeds in the 1st harvest than the Westerwold species. Westerwold is thus considered a better fir for fermented silage.

(c) www.herrinhs.org
Red clover
Red clover has a high amount of crude protein and is highly palatable, but may be difficult to store by fermentation. Is has less fibre than hay. Due to the high palatability animals eat red clover more than grass-based silage even though the clover had a lower D-value and less fibre. Red clover feed increases milk yield and the amount of unsaturated fats in the milk. It is recommended to be used together with grass- or hay-based silage. Note that red clovers are high in potassium, so grazing animals on pastures with plenty of red clover may lead to grass staggers.

(c) www.innerpath.com.au
Lucerne / alfalfa
The queen of crops, lucerne is the most important silage ingredient in the world. It is used as silage, dried and as pastures. Lucerne is very high in crude protein, so it should be mixed with hay or grass to bring the CP down to a reasonable level. Lucerne also has lower fibre content and lower digestibility than hay, which are compensated by higher palatability. Compared to red clover, lucerne affects milk yield as well as or even better than red clover.

The gross energy of straw is high, but due to extremely low digestibility it yields practically no energy to the cow. Straw in a whole crop silage has more energy and higher digestibility than that of a fully ripened crops.  It is mostly used to increase the fibre content especially for dry cows and heifers. Straw is also very good for beed cattle fed with high amounts of concentrate, and for suckler cows. Straw is not necessary for cattle fed with hay- or grass-based silage.

whole crop silage
(c) takakita-net.co.jp
Whole crop silage
Wheat and barley are the most used species, because they have the highest palatability. Oat has a lower digestibility, and is used mostly for dry cows and heifers. Rye is palatable only if harvested early, but even then it is usually too expensive to be used for cattle feed. Whole crop silage can be used alone or together with grass- or hay-silage to offset the high crude protein content of the silage.

Harvesting of whole crop silage must be considered carefully. If it is harvested too early, the feed has low dry matter content and low digestibility, and may be difficult to preserve by fermentation. Late harvesting causes losses due to shedding, and the feed does not ferment. In addition to harvesting, the crops used and their properties impact the quality of whole crop silage.

Peas. (c) TerraLink
Peas have plenty of crude protein, and are best used in together with whole crop silage. Growing pea together with grains reduces the need for nitrogenous fertilizers and increases the protein content and digestibility of the silage. However, shedding losses may be noticeable, and peas are difficult to preserve by fermentation. Peas are annual plants and the seeds may be expensive, but if the harvesting is done well, the yield is high.

(c) Suomen Rehu
Commercial concentrates
Concentrate is used to increase the amount of one or more nutrient in the feed mix. Energy feeds and protein feeds are the most common ones used, but minerals and vitamins are also often part of concentrates. The impact of increased protein and energy on milk yield is limited, and can be minimal if the D-value of roughage is high. Too high amounts of concentrate reduce the amount og roughage eated, which again reduces the amount of fibre gained, and leads to bloat and acidosis. A commercial "full concentrate" has no grains, and contains usually 30-45 % CP and 10-11 ME MJ. A "semi-concentrate", which includes grains, has 23-30 % CP and 12-13 ME MJ.

More information

Dairy animal nutrition by Penn State University

Other posts in this blog about animal nutrition:
Feeding beef cattle
Feeding poulty
Legumes, rapeseed and grains in pig production
Determining feed digestibility
Value, quality and preservation of silage


Saturday, 26 April 2014

Synthesis and composition of cow's milk

Alveoles of the udder.
(c) Sjaastad et al., Scanvetpress
Cow's udder consists of four independent mammary glands, which each have one teat and one opening. Milk is produced in alveoles. Each alveole contains 0,01 ml of milk, which is secreted from the epithelial cells. Once the alveole is full, myoepithelial cells outside the alveole constrict and push the milk into the milk duct, and milk chamber. All in all the udder of an adult cow weighs 10-30 kg, and has 5-20 milk ducts which branch out to smaller milk ducts. The ducts end in lobes, which are filled with alveoles in smaller lobules. The udder needs 400-500 liters of blood to absorb enough nutrients for a kilogram of milk.

Udder development (mammogenesis) begins during the embryonic development. During the first three months of gestation the embryo develops teat channels, milk chambers and initial mammary glands. Actual ducts, teats and fat pads to protexct the udder are formed before the calf is born. From birth until 3 months of age the udder develops isometrically, i.e. as fast as other tissues. Between 3 months and 1 year of age the udder develops allometrically, i.e. faster than other tissues, and then returns to isometric development. Too heavy feeding before puberty decreases the amount of secreting tissue and reduces milk yield.

When the heifer is in gestation, the udder development continues. The ducts grow and increase during the first 3 months, and later the portion of the secreting tissue increases. 10 % of secreting milk ducts develop only after calving, which causes the lactation peak about 2 months after calving. After the peak the number of alveoles begins to decrease. The decrease is faster if the cow gets pregnant again.

Lactogenesis is the actual synthesis of milk, which cab begin once the cow has reached puberty. 0-4 days before calving the amount of progesterone in the body decreases but prolactin and glucocorticoids increase. This signals the body to start producing alpha-lactalbumin, a gene which starts the lactose synthesis in the Golgi apparatus. Lactose changes the osmotic pressure of the udder and thus draws water into the udder. The maintenance of milk production is called galactopoiesis. Prolactin is a hormone which is secreted during milking, and increases the metabolism of the epithelial cells of the milk alveoles. Insulin-like growth factor IGF-1 maintains galactopoiesis and increases milk yield. This is why some countries allow synthetic IGF-1 called BST, Bovine Somatotropin, be injected into the cows to force their metabolism to produce even more milk.

The components of milk are synthetized in different parts of the mammary gland. Lactosis is synthesised in the Golgi apparatus from glucose and VFAs in the blood stream. Lactose is the main component which determines the amount of milk produced. Fats of the milk are synthesized in the cytoplasm from VFAs other fatty acids. Milk protein is synthesised in ribosomes from amino acids absorbed from the blood stream. Minerals and vitamins are secreted into milk directly from the blood stream.

Composition of milk

On average, 87 % of milk is water, 3,25 % is protein and 3,9 % is fat. The rest is a variaety of minerals, trace minerals, vitamins, acids, enzymes and gases. 95 % of the nitrogen in milk is in proteins, the rest is in urea, ammonia and creatine. The composition is affected by several factors:
  • Breed: Jersey cows produce milk with higher fat and protein content than ayrshires or Holsteins.
  • Phase of lactation: fat and protein content increase towards the end of the lactation. Also right after calving the colostrum is rich in fat and protein,
  • Milking technique: Incomplete milking leaves the fatty milk inside the udder. and long intervals (over 16 hrs) between milkings decrease milk yield.
  • Udder health: during mastitis there's more Cl and Na in the udder, and less K.
  • Somatic cell count: the higher the count, the lower quality the milk has. Excellent milk has less than 250 000 somatic cells in a 3 mth average geometrical calculation.
  • Proportions of the VFAs released from the rumen: butyric acid increases fat synthesis, acetic acids maintains it and propionic acid decreases it while increaseing protein synthesis.

Effect of diet on milk production
(c) Babcock Institute e-learning
Feeding is a very important factor in determining the milk yield and the composition of milk. Generall,  high energy intake decreases fat and protein synthesis due to the increase of propionic acid synthesis. However, if the high energy content is due to highly digestable roughage, then fats and proteins in the milk may be increased. Highed energy content generally increases protein synthesis due to increased energy available for the rumen microbes.
  • Adding silage or concentrate: increases protein synthesis, decreases fat synthesis
  • Adding protein to the food:Same as above, increases the amount of urea in the milk.
  • Adding fats to the feed: decreases protein synthesis, may decrease fat synthesis
  • Better digestibility of the roughage and limiting the fermentation of the roughage: increases fatty acid and protein synthesis. Protein synthesis is increased more than fatty acid synthesis.
  • Using hay or legumes in grass-based roughage decreases protein synthesis. However, especially red clover increases milk yield (~1,4 kg / d)  and protein & fat synthesis, but due to the increased yield the percentage of both fat  and protein in the milk is decreased (-1,5 g/kg and -0,4 g/kg, respectively).

Grazing on a pasture also affects the composition of milk. During grazing season the amount of unsaturated fatty acids in the milk increase due to the composition of grass versus fermented roughage. The relatively high fat content of grass disturbs the biohydrogenation of the rumen, which leads to less fatty acids being saturated. Protein content of milk decreases during grazing. In countries where cows graze only during summer there is clear annual changes in the composition of milk. Both protein and fat content decrease during the summer, and increase again once the cows are moved back indoors.

More information

Dairy Education Series by Professor H. Douglas Goff, Dairy Science and Technology Education Series, University of Guelph, Canada.

Dairy Essentials by Babcock Institute for International Dairy Research and Development, University of Wisconsin-Madison, USA.

(c) Zweber family farms

Tuesday, 22 April 2014

Basics on animal genetic resources

Farm animal genetic resources, or simply AGR, refers to the genetic material we currently have in live animals and frozen in sperm banks, and which is of economical, cultural and scientific importance. Genetic diversity, both within and between breeds, is vital. Diversity
  • helps the animals to adapt to their environment
  • is the basis for animal breeding
  • allows adaptation to changes and new breeding targets
  • prevents inbreeding depression
  • keeps the frequency of harmful allelels low.
Domestication has already reduced genetic diversity in farm animal species (and artifical insemination has reduced it even further). Domestication is the process by which captive animals adapt to man and the environment provided, and it is achieved through genetic changes. Genetic factors affecting domestication are inbreeding, genetic drift and selection.  

Impact of domestication on milk yield of dairy cows.
Inbreeding is necessary when selecting a desired trait, but it reduces heterozygosity and thus diversity, although it does not affect allele frequencies. Genetic drift causes alleles to become fixed or deleted randomly, and its direction cannot be estimated. Selection, both natural and artificial, has altered the fitness of certain traits. For example, in domestication species the distance to flee and fearfulness have been decreased, even though they are vital for a wild animal. These are both behavioral changes. Physiological changes involve changes in hormone levels, reproduction cycle and production traits (e.g. the increased milk yield of cows and the all-year farrowing of sows). Morphological changes have also occurred, as animals have grown larger and developed colors unseen in the wild (especially white). For an example of scientific study on domestication, see Giuffra et. al (2000) or Kantanen et al. (1999).

Studying domestication

Domestication can be studied in several methods. Paternal transmission of Y-chromosome shows traits have developed from sire to offspring. It's counterpart is the study of mtDNA (mitochondrial DNA), which is always inherited from the dam to all her offspring. mtDNA haplotypes are sequenced and aligned, and dendrograms or cladograms are drawn based on the multiple-sequence-alignment (MSA) results. The haplotypes can be further divided into groups, which helps to draw a network. For more information visit the blog The Genealogical World of Phylogenetic Networks and their post on interpreting rooted networks.

The most common method by far is still studying microsatellite markers. Usually 20-30 microsatellites are studied, but FAO has published recommendations for each animal species (FAO). Polymorphic loci with 4 or more alleles are recommended to eliminate false positives by identical-by-state -alleles. Unlike mtDNA and Y-chromosome, microsatellites are inherited from both parents to all offspring. Genotyping and aligning SNP-markers is similar to microsatellies, but requires the use of thousands or hundreds of thousands of SNP-markers. Microchip arrays are readily available for several species for SNP-analysis.

Practical application of genetic domestication studies.
(c) ILRI 2006

Studying ancestral DNA is tedious, but can yield valuable information on extinct species. Ancestral DNA can be thousands or tens of thousands of years old. It is collected either from animal remains such as fossils, teeth, wools, hides or bone pieces, or from the ground ("dirty DNA"). Ancestral DNA also helps to chart the spread of different animal and plant species and to determine temporal changes in their genetics. The problem with ancestral DNA is that the concentration of desired DNA is often low, while the concentration of microbial DNA is high. Contamination risk is very high indeed. Sterile environment must be maintained whenever possible when working with ancestral DNA. Another problem is that the ancestral DNA has been fragmented, and many chemical bonds have been broken. C > T and G > A mutations in PCR are common due to deamination. The results must be confirmed in several independent laboratories. In addition, all results derived from ancestral DNA must fit in to earlier context.

Retroviruses have been used as a study method on sheep. Retroviruses are viruses, which insert their RNA to the sheep's system, and with reverse-transcriptase produce DNA from the RNA. The viral DNA is then integrated as a part of the sheep's own genome. The viral DNAs which have infected germline cells are hereditary, and thus provide a tool for studying the evolution of sheep. The original virus infections happened 5-7 million years ago, and have continued to branch even during the last 10 000 years. Studies of retrovirus-DNA has shown that originally all sheep in Europe were used for meat production. A meat-and-wool producing breed was introduced later,  and replaced the old breed nearly completely. Still existing breeds originating from the first migration are Soay sheep, Gutesheep and Finnsheep.  

Determining the level of endangerment and the value of a breed

There are thousands of animal breeds in the world. FAO's DAD-IS -information system classifies breeds into four categories:
  • local breeds, which exists in one country or area only
  • transboundary breeds, which exists in several countries
  • regional transboundary breeds, which exists in several countries but only in one continent
  • international transboundary breeds, which exists in many continents.
Each breed is also classified based on the level of endangerment. There are five levels, which are determined by the population size and number of breeding males and females. Other classification systems also consider the direction of population size (growing or decreasing), the purity of the species and the number of populations (e.g. herds). FAO's five levels are
  1. Extinct
  2. Critical (with or without a conservation program)
  3. Endangered
  4. Not at risk
  5. No information on the population size.
Currently DAD-IS lists (among others) 3093 cattle breeds, 2558 chicken breeds/lines and 1283 pig breeds. Altogether there are 14544 breeds listed for 38 animal species. Of the listed local cattle breeds, 181 are extinct (209 in 2006) and only 399 are not at risk. For pigs there are 110 extinct breeds (140 in 2006) and 206 not at risk. The numbers from the year 2006 are larger, probably due to renewal of the concept of breed or improved methods of separating breeds and collecting information. Below are a few examples of the tables created from DAD-IS system, showing the status of cattle, pig and sheep breeds in different regions.

The level of endangerment is the likelihood of the breed going extinct in the current circumstances within a certain time period. It can be used to estimate how much there is time to save a breed. The level depends on demographic factors (population size and its changes) and genetic variation. Genetic variation is calculated from effective population size Ne, which again is deduced from the change of inbreeding (Ne = 1/2 ΔF). Growth factor can be calculated from r = anti-log (( logN2–logN1) / t ), where N1 and N2 are the population size at two different measurements (generation 1 and 1+n), and t is time in years. The growth factor depends on animal births and deaths, cullings, changes in market prices and agricultural politics and on epidemics.

Method of estimation impacts the value of  ΔF. Pedigree-based studies give consistently lower estimated of ΔF than SNP-based evaluation. In one study, pedigree analysis found that 15 % of animals had F > 6.25 %, while in an SNP-study the percentage was 25 %. For pairwise kinship coefficients both methods are equally reliable for close relatives. For more distant relatives the pedigree analysis gives higher estimates than SNP-analysis. (Li et al. 2011)

However, Ne, ΔF and the growth factor are only single meters. To estimate the value of a breed for genetic conservation requires a more holistic approach. In addition to the meters mentioned before, the breed value depends on several factors. The factors, and examples related to them, are listed below.
  •  its ability to adapt to a certain environment (Yakutian cattle, goat breeds in arid African countries)
  • economically important traits (the excellent cheese-making qualities of the milk of Finncattle)
  • unique traits (breed-specific mutations, alleles and gene combinations)
  • cultural heritage, historical value (Yakutian cattle)
  • unique genetics.
One must remember that the breed must be able to cope even in the future, and to continue being useful for the herders. It is not a viable option to maintain breeds which cannot survive for example after global warming or if their surroundings change due to industrialization. 

Yakutian cattle (c) EPFL / Anu Osva

Thursday, 13 March 2014

Genetic analysis

XKCD's take on genetic analysis.
Genetic analysis may sound complicated, but it relies on very simple principles of heredity, statistics and probabilities. Sometimes the genotypes are not known at all, and we need to look at pedigrees to determine models of heritance. Here some basic principles of genetic analysis are discussed. Later on a post on bioinformatics will make a deeper analysis on how to analyze known genes an genomes.

Calculating probabilities

Inheritance, or which genes each parent passes on to the offspring, is always partially random. This is why probabilities are important. For example if we know the absolute frequency, i.e. the number, of a certain allele in a sample, we can deduce the probability of one randomly picked individual having that particular allele. Set the absolute frequency of the allele a to 118. Now we know that there are 118 a alleles in our sample of 500 chromosomes. The probability of a random individual to have a is simply the amount of a divided by the number of all possible alleles: 118 / 500 = 0,236. This is also the relative frequency of a.

So, the probability of A, whatever A is, is P(A) = the number of favorable results / the number of all possibilities. The probability of A's complement, of A not happening, is 1- P(A). Combinations of two or more independent variables are calculated as follows: P(A and B) = P(A) * P(B) and P(A or B) = P(A) + P(B). An example of a complement: if a chromosome has the allele a, it cannot have the allele A. Only either one or the other (barring some rare genetic mutations, which are not considered here).

The probability of two separate outcomes depends on if the outcomes are related or not. A union means that either (A or B) or (A and B) happen. For independent variables the union would be P(A) + P(B) - P(A u B). For dependent variables the union is zero: if A has happened, B cannot happen, or vice versa.

To ease your stress, here is a cute animal.
The conditional probability, the probability of A happening when we know that B has already happened and A and B are dependent, is (P(A) * P(B)) / P(B). This is often marked as P(A|B).

Permutations and combinations deal with the order of several possibilities. Permutations are used when the order of the events is important, and is counted as n!. Combinations take all orders into consideration, and are counted as n! / (r! (n-r!)).

Binomial probability is a bit more complex. It is used when we want to determine the probability of getting exactly r favorable results, each with the probability of p, out of n repeats. The magic word here is exactly - when that is used in an exercise, think of binomials.

More on probabilities:
For the mathematically gifted: Wikipedia 
Cut the Knot (also covers Bayesian methods)


Statistics have been covered before in a post named Variance Analysis (one of the most popular posts in this blog!), so I'll just remind you of the formulae we will need when doing genetic analysis.

There are thousands of helpful websites you can look up for more information on statistics. Here are just a few:

Mendelian genetics by Phillip McLean


Right, let's get down to the real deal and do some analysis! The examples are from University lecture materials for course in genetic material, but I unfortunately cannot share the entire material due to copyright restrictions and a language barrier - the materials are not in English :)

Example 1. Parents are heterozygotic concerning their eye color. They both have the allele s for blue eyes and the allele S for brown eyes. Calculate the probability that they will have 
a) a child with blue eyes  b) five children with blue eyes.

From the way the alleles are written we see that S is the dominant allele. The genotypes of the parents are Ss and Ss, so the possible genotypes of their children are 

Now we can see that 3/4 of the offspring have the dominant allele, so only 1/4 has blue eyes (homozygote ss). Therefore the P(a child has blue eyes) is 1/4 = 25 %. The probability for each subsequent child is similar, so P (five of five children have blue eyes) is 0,25 * 0,25 * 0,25 * 0,25 * 0,25 = 0,000977.

Example 2. 32 % of people infected with a rare illness have mutation A, and 16 % have mutation B. 10 % of those infected have both A and B. Calculate the probability of randomly selected person to have at least one of the mutations? 

The selected person must now have either A or B or both. What is needed is the union of A and B: P(A) + P(B) - P(A u B) = 0,32 + 0,16 - 0,1 = 0,38. 38 % have either A, B or both mutations. Note that the union P(A u B) is NOT A*B in this case, but it is given as 10 %.

Example 3. There are 2 boys and 5 girls in a family. In how many different sequences could the children have been born? 

Because the order is not important, we'll use combinations: n over r, i.e. n! / r!(n-r)!. The n now is 7, the number of all kids. The r can be either 2 or 5: both give the same result. If r = 2, then 
7! / 2!(7-2)! = 7! / 2! 5! = 5040 / 240 = 21.

Example 4. Parents are heterozygotes concerning a rare recessive illness. Calculate the probability that out of three children
a) all are healthy
b) two are ill
c) at least two are ill.

a) Recessive heterozygotes produce 25 % of recessive homozygotic alleles (see example 1). The probability for each child to be healthy is thus 1 - 0,25. P(all are healthy) = 0,753 = 0,422.

b) Two out of three must be ill, so we need the binomial distribution. Now r = 2, n = 3 and p = 0,25. The first factorial, n over r, gives 3!/2!(3-2)! = 3. Continuing from there we have 3 * 0,252 * (1-0,25)3-2 =3 * 0,0625 * 0,75 = 0,141.
c) If at least two must be ill, then the probability is P(two are ill) + P(three are ill). The first part is calculated like in part b: 3 * 0,252 * (1-0,25)3-2 =3 * 0,0625 * 0,75 = 0,141. P(three are ill) is simply 0,253 = 0,015625. So P(at least two are ill) = 0,141 + 0,015625 = 0,156. 

Example 5. The penetrance of a certain illness varies between genotypes. The penetrances are 0,01 for AA, 0,05 for Aa and 0,5 for aa. In a population the allele frequencies are  f(a) = 0,05 and f(A) = 0,95. The population is in Hardy-Weinberg equilibrium. Count the prevalence of the illness in the whole population.

Whoa, lots of terms here! Penetrance is the probability of expressing a certain trait. Here the trait is the illness. H-W balance means that in an ideal population, where q and p are the relative frequences of a alleles, there are p2 dominant homozygotes, 2pq heterozygotes and q2 recessive homozygotes. What are they actually askingfor is the prevalence, i.e. the probability of a random member of the population to have the illness.

First we need to calculate the frequencies of the genotypes in the population. With the H-W equilibrium and the given frequencies we know that p = 0,05 and q = 0,95. Now the genotype frequencies are  
AA  = dominant homozygotes = p2 = 0,952 = 0,9025
Aa = heterozygotes = 2pq = 2*0,95*0,05 = 0,095
aa = recessive homozygotes = q2 = 0,052 = 0,0025

Now each genotype has its own probability of actually expressing the illness. To make it easier to understand we can build a table of  a tree of probabilities:

So now the random person we select has a 0,9025 % chance of having the genotype AA, and then 0,01 % chance of being sick.  Note that these variables are independent (a person can only have one genotype and be either sick or healthy). The possibility of expressing the illness is thus P(has a certain genotype) * P(is sick).

P(is sick) = P (AA and sick) + (P Aa and sick) + P(aa and sick) = P(0,9025 * 0,01) + P(0,095*0,05) + P(0,0025*0,5) = 0,009025 + 0,00475 + 0,00125 = 0,015.

Example 6. The observed genotype frequencies in a population are f(AA) = 31, f(Aa) = 89 and f(aa) = 122). Is the population in Hardy-Weinberg equilibrium?

If we were mean about it, we'd say no because no actual population is ever in H-W equilibrium. However in an exam we'd get 0 points for that, so let's calculate this. Now we need to use x2 or khi squared test. We already have the observed frequencies. Now we need the expected frequencies, i.e. the frequencies if the population was in H-W equilibrium.

To get there we calculate the allele frequencies in the population. There are 31 creatures with two A alleles and 89 with one. In total we then have 2 * 31 + 89 = 151 A-alleles. The recessive allele a is calculated similarly: 2*122 + 89 = 333. In total there are 151 + 333 alleles, so the relative frequencies are 151/484 = 0,312 for A and 1-0,312 = 0,688 for a. So now p = 0,312 and q = 0,688.The expected absolute genotype frequencies are now
AA  = p2 = 0,3122 * 242 (the size of population) = 23.56
Aa = 2pq = 2 * 0.312 * 0.688 * 242 = 103.89
aa = q2 = 0.6882 * 242 = 114.55

We need to calculate the chi squared test value using the formula

To make it easier, let's put our values to a table. Then we can use the formula and calculate the x2 test variable.

The value 4.97 is not the answer. Remember the question: is the population in H-W equilibrium? The answer hides in a x2 distribution table. To use that we need degrees of freedom (df), which in an chi squared good of fit test is the number of classes minus 1. Here we have three classes (three genotypes), so our df = 3-2 = 1. Now we look at a x2 distribution table such as this.

With df = 2 we find that our  x2-value, 4.97, goes between 4.605 and 5.991. The corresponding P-values are 0.10 and 0.05. So our p is 0.1 - 0.05. Without going too far into interpreting p-values we can just note that it is higher than 0.05 which means that we abadon the hypothesis 0 (the population is in H-W equilibrium). P > 0,05 shows us that the population is NOT in H-W equilibrium.