3) A kangaroo with a long tail and brown fur and a kangaroo with a short tail an

Question

3) A kangaroo with a long tail and brown fur and a kangaroo with a short tail and black fur mate. The joey (baby kangaroo) has a short tail and black fur. Note that long tail (S) is dominant to short tail (s) and brown fur (B) is dominant to black fu (b) What is the genotype of each parent? long tail and brown fur genotype - short tail and black fur genotypes a. b. What genotype are the gametes made by the parent with the short tail and black fur c. What is the genotype of the joey 4. In the following cross, the genes are inherited independently except for C and D, which are tightly linked and show zero recombination: A/ab/b /c; D/D; e/e;F/Fx a/a; B/B; C/C;d/d; E/E;f/f. In the Fz, what proportion of individuals will be: a) pure breeding? b) heterozygous for all loci? c) homozygous recessive for all loci? e) phenotypically like either parent? f phenotypically unlike either parent? g) genotypically like either parent? h) unlike either parent? 5. Identify the parental genotypes and draw a recombination map of the v, cv, and loci Recombinant for loci v and cv and ct cv and ct 580 592 45 v' ·CV·ct v cv ct vCVct

Explanation / Answer

Now we know-

long tail (S) is dominant upon short tail (s)

and brown (B) is dominant upon black (b)

Now joey the baby kangaroo has short tail black fur. It will only happen when there wouldd be no dominant form of allele that is no "S" or "B".

so the genotype of baby kangaroo(joey) will be- ssbb

Now comming to the parents genotype-

short tail black fur will be- ssbb,

and long tail brown fur has to be- SsBb, only then a baby kangaroo with ssbb genotype can be formed.

Now parent with short tail and black fur (ssbb) will produce only one type of gamete that is - "sb"

and parent with long tail and brown fur (SsBb) will produce 4 gametes- SB, Sb, sB and sb.

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