LO21 Calculate genotypes and phenotypes in sex-linked, sex-limited and sex- influenced traits The presence of a beard on some goats is determined by an autosomal gene that is dominant in males and recessive in females. If we cross a beardless male with a bearded female, what is the probability of obtaining a bearded female in the offspring? 100% C 50% 0% 25% O 75%
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- Sixteen percent (16%) of the population is homozygous recessive for a trait. What are the frequencies of the dominant and recessive alleles of that trait?A1) The allele for widows peak (H) is dominant for the allele for no widows peak (h). At a different gene locus, the allele for hitchhikers thumb (D) is dominant to the allele for non-hitchhikers thumb (d). A man is heterozygoous for the traits and marries a woman who has no widows peak and is heterozygous for hitchhikers thumb. What fraction of this couple's children should have no widow's peak and no hitchhiker's thumb? 1/4 1/8 1/2 3/8 3/4 A2)Using the forked line method, determine the phenotypic ratio of a trihybrid cross between AaBbCc X AaBBCC. What proportion of offspring will have the dominant phenotype for A, B, and C? 1/8 1/4 1/16 3/4 1/2 A3) How many different offspring genotypes are expected in a trihybrid cross between parents heterozygous for all three traits when the traits behave in a dominant and recessive pattern? How many phenotypes? 8 genotypes; 27 phenotypes 27 genotypes; 8 phenotypes 16 genotypes; 64 phenotypes 64 genotypes; 16 phenotypesQ1: What is the probability that a child with one parent who has an autosomal dominant disorder will inherit the disease? Q2: Why are there no carriers with a dominant genetic disorder? Q3: Because dominant genetic disorders are rare, it is extremely rare for both parents to have the condition (genotype Aa). Draw a Punnett square with two Aa parents. What proportion of the offspring would have the disorder? What proportion would be normal?
- Punnet square problems A=Codominant; B=Codominant; O=Recessive Mary is homozygous for type A blood. Steve is homozygous for type O blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Mary and Steve have a son, Brad. Brad’s wife, Samantha is heterozygous for type B blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Stella loves roses and decides to cross her red rose with her white rose. All of the resulting offspring of this cross are pink roses. What can you say about the red and white alleles as a result of this cross? Stella decides to cross two of the pink roses. What are the possible genotypes and phenotypes of the offspring and the probabilities of each? DNA replication, Transcription and Translation problems It is S phase of the cell cycle, and time to replicate the cell’s DNA. Using the following strand of DNA…A1) The allele for widows peak (H) is dominant for the allele for no widows peak (h). At a different gene locus, the allele for hitchhikers thumb (D) is dominant to the allele for non-hitchhikers thumb (d). A man is heterozygoous for the traits and marries a woman who has no widows peak and is heterozygous for hitchhikers thumb. What fraction of this couples children should have a widow's peak and a hitchhiker's thumb? 3/8 1/4 1/2 3/4 0 A2)The allele for widows peak (H) is dominant for the allele for no widows peak (h). At a different gene locus, the allele for hitchhikers thumb (D) is dominant to the allele for non-hitchhikers thumb (d). A man is heterozygoous for the traits and marries a woman who has no widows peak and has no hitchhikers thumb. What fraction of this couple's children should have no widow's peak and a hitchhiker's thumb? 3/8 0 1/4 1/2 1/8ASAP PLEASE In Talons, the feral style (S) is dominant over the gladiator style (s), and green eyes (G) are dominant over yellow (g). If a Talon that is heterozygous for both traits is crossed with a Talon that is homozygous recessive for both traits, what is the predicted genotype of the offspring? A)SsGg and Ssgg B)SsGg only C)SsGg, ssgg, ssGg, Ssgg D)ssGg and ssgg only
- Q. 1: In rats, the following genotypes of two independently assorting autosomal genes determine coat color:A–B– (gray)A–bb (yellow)aaB– (black)aabb (cream)A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles.However, the cc genotype results in albino rats regardless of the A and B alleles present.Determine the F1 phenotypic ratio of the following crosses:(a) AAbbCC × aaBBcc(b) AaBBCC × AABbcc(c) AaBbCc × AaBbcc(d) AaBBCc × AaBBCc(e) AABbCc × AABbcc The above sub-part of question is already solved Q. 2: Given the inheritance pattern of coat color in rats described in Q 3, predict the genotypeand phenotype of the parents who produced the following offspring:(a) 9/16 gray: 3/16 yellow: 3/16 black: 1/16 cream(b) 9/16 gray: 3/16 yellow: 4/16 albino(c) 27/64 gray: 16/64 albino: 9/64 yellow: 9/64 black: 3/64 cream(d) 3/8 black: 3/8 cream: 2/8 albino(e) 3/8 black:…Baldness is a sex linked trait. What parental genotypes could produce a bald woman? Use H for normal hair And h for bald Draw 6 punnet square 3 for female and 3 for maleBbRrppMm X bbRrPpMM: probability of producing an individual that is dominant for B & M and recessive for R and P?
- A phenotypically normal couple has had one normal child and one child with cystic fibrosis, an autosomal recessive disease. What are the genotypes of the parents? What is the chance that their next child will be a carrier of cystic fibrosis?Many recessive traits are very rare in a population, but cystic fibrosis does not follow this pattern. In some populations, the frequency of carriers for cystic fibrosis is 1/35. Does this mean that cystic fibrosis is not more common in inbred mating, such as those between first cousins? Why or why not?Q. 3: In rats, the following genotypes of two independently assorting autosomal genes determine coat color A–B– (gray A–bb (yellow aaB– (black aabb (cream A third gene pair on a separate autosome determines whether or not any color will be produced The CC and Cc genotypes allow color according to the expression of the A and B alleles However, the cc genotype results in albino rats regardless of the A and B alleles present Determine the F1 phenotypic ratio of the following crosses (a) AAbbCC × aaBBc (b) AaBBCC × AABbc (c) AaBbCc × AaBbc (d) AaBBCc × AaBBC (e) AABbCc × AABbc Q. 4: Given the inheritance pattern of coat color in rats described in Q 3, predict the genotyp and phenotype of the parents who produced the following offspring (a) 9/16 gray: 3/16 yellow: 3/16 black: 1/16 crea (b) 9/16 gray: 3/16 yellow: 4/16 albin (c) 27/64 gray: 16/64 albino: 9/64 yellow: 9/64 black: 3/64 cream (d) 3/8 black: 3/8 cream: 2/8 albino (e) 3/8 black: 4/8 albino: 1/8 creamAn individual with a form of red-green color blindness processes a genetically inherited trait that makes it difficult to distinguish red and green color hues. Red-green color blindness tends to skin generations and it is found much more often in men than in women. If a man who was normal and a woman who is a carrier for this form of colorblindness have a child, when the probability that the child will be red-green colorbind is: -0% -25% -50% -75% -100%