An Sci 361 1st Edition Lecture 14 Outline of Last Lecture I. InbreedingII. SelfingIII. Consequences of Increased Homozygositya. Prepotencyb. Expressionc. Inbreeding DepressionIV. Inbreeding Depressiona. MeasuringOutline of Current Lecture I. Testing for recessivesa. ExamplesII. Homozygous Dominant vs. Heterozygousa. Biochemical Testingb. DNA Testingi. BLADii. Mulefootiii. DUMPSiv. CVMv. Citrullanaemiac. Test MatingsIII. Pr[D] with different types of matesa. Homozygous Dominantb. Known Carrierc. Homozygous Recessived. Individual’s Daughterse. Mate’s Chosen randomly from PopulationIV. Characteristics of Different Mating Testsa. Homozygous Recessive MatesThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.b. Carrier (heterozygous) matesc. Daughtersd. Random mates from the populationCurrent LectureTesting for Recessives- Selection for a particular phenotype is easy when each genotype results in a different phenotype, e.g. traits that exhibit no dominance, partial dominance, or over dominance.o Examples: Coat color in Shorthorn cattle:- RR: red- Rr: roan- rr: white Ear length in goats:- LL: long- Ll: short- ll: earless Coat color in Shetland sheepdogs:- MM: white merle (mostly white)- Mm: merle (dappled with color)- mm: normal color (can vary)- With complete dominance, selection for the phenotype of the homozygous recessive genotype is easy.o Examples: Red and White Holstein cattle (bb) Red Angus cattle (bb) Droop-eared pigs (ee) Single-combed chickens (rr)- However, selection for the dominant phenotype with complete dominance is less certain.o Examples: Black and white Holstein cattle (BB, Bb) Black Angus cattle (BB, Bb) Erect-eared pigs (EE, Ee) Rose-combed chickens (RR, Rr) Homozygous normal individuals versus heterozygous carriers of a recessive disorder - You wish to select homozygous dominants and cull heterozygotes. How do you tell the difference?Differentiating between homozygous Dominant and Heterozygous individuals- 1. Biochemical testingo Example: citrullinaemia in cattle - an enzyme (argininosuccinate synthetase, ASS) deficiency results in failure of the conversion of citrulline to urea in the urea cycle, a build up of ammonia, and death of the calf. AA: normal cattle – high levels of ASS Aa: normal cattle – ½ the level of ASS of AA cattle aa: affected cattle (calves die) – no ASS o An assay for ASS will differentiate AA from Aa individuals.- 2. DNA testing - direct identification of each alleleo Example: The causative mutation for many recessive genetic defects in livestock are now know, and DNA testing can identify carriers. - Bovine Leucocyte Adhesion Deficiency: BLAD in Holstein cattle is an autosomal recessive congenital disease characterized by recurrent bacterial infections, delayed wound healing and stunted growth, and is also associated with persistent marked neutrophilia.- Syndactyly (Mulefoot): Syndactyly refers to the fusion of the two toes of the foot. Caused by a recessive gene, mulefoot most often affects the front feet.- Deficiency of Uridine Monophosphate Synthase (DUMPS): monogenic autosomal recessive disorder in cattle, resulting in early embryonic death of homozygous offspring.- Complex Vertebral Malformation (CVM): The most noticeable characteristics of CVM-affected calves are malformed legs with flexed and rigid pasterns, a shortened neck and an abnormal curvature to the spine. Most of the affected calves are reabsorbed as embryos or aborted as fetuses; the remaining pregnancies result in a stillborn calf. - Citrullinaemia: an enzyme (argininosuccinate synthetase, ASS) deficiency results in failure of the conversion of citrulline to urea in the urea cycle- 3. Test matings - mate the individual being tested o If at least one homozygous recessive progeny is produced, the individual is heterozygous.o If no homozygous recessive progeny are produced, probability theory can be used to give us a level of “confidence” in our results.- Confidence in the test matingo If the individual tested is a heterozygote (Bb), what is the probability he or she will be detectedo Prob. of detecting Bb = Prob. of at least 1 bb (red) progeny = Pr[D]o Examples: If a Bb bull is mated to 1 Bb cow and the cow gives birth to 1 calf, what is Pr[D]?- Bb bull x Bb cow ® 0.25 BB, 0.50 Bb, 0.25 bb® 0.75 black, 0.25 red- Pr[D] = prob. of a red calf = 0.25 If the bull is a carrier, this mating test will detect him as a carrier only 25%of the time. If a black calf is born, we are only 25% “confident” that the bull is not a carrier (BB).  If a Bb bull is mated to 2 Bb cows and each cow gives birth to 1 calf, what is Pr[D]?- Bb bull x Bb cow ® .25 BB, .50 Bb, .25 bb ® .75 black, .25 red- Pr[D] = prob.[1st calf red and 2nd calf black] + prob.[1st calf black and 2nd calf red] + prob.[1st calf red and 2nd calf red]= [0.25 x 0.75] + [0.75 x 0.25] + [0.25 x 0.25] = 0.1875 + 0.1875 + 0.0625 = .4375- If the bull is a carrier, this mating test will detect him as a carrier 43.75% of the time.- If 2 black calves are born, we are 43.75% “confident” that the bull is not a carrier (BB).  If a Bb bull is mated to 16 Bb cows and each cow produces one calf, what is Pr[D]?- Pr[D] = Prob. of at least 1 bb (red) progeny = 1 - probability of all progeny BB or Bb (black)- Pr[D] = 1 - (1st calf black x 2nd calf black x … x 16th calf black) = 1 - (0.75 x 0.75 x … x 0.75) = 1 - (0.75)16 = 1 - 0.0100226 = .99 If the bull is a carrier, this mating test will detect him as a carrier 99% of the time. If 16 black calves are born, we are 99% “confident” that the bull is not a carrier (BB).Pr[D] with Different Types of Mates Used in Tests (1 offspring per mate, n = number of mates)- 1. Homozygous dominant (BB)o Bb x BB ® 0.50 BB, 0.50 Bb ® 1.0 Blacko Pr[D] = 1 - P(all black) = 1 - 1n = 0- 2. Known carrier (Bb)o Bb x Bb ® 0.25 BB, 0.50 Bb, 0.25 bb ® 0.75 black, 0.25 redo Pr[D] = 1 - Pr(all black) = 1 - (0.75)n - 3. Homozygous recessive (bb)o Bb x bb ® 0.50 Bb, 0.50 bb ® 0.50 black, 0.50 redo Pr[D] = 1 - Pr(all black) = 1 - (0.5)n- 4. Individual's daughters (assume dams of the daughters were BB)o Bb x BB ® 0.50 BB, 0.50 Bbo Bb x (0.50 BB, 0.50 Bb) ® 0.375 BB, 0.50 Bb, 0.125 bb ® 0.875 black, 0.125 redSire frequenciesDaughter0.75 B 0.25 b0.5 B 0.375 BB 0.125 Bb0.5 b 0.375 Bb